Problem: What is the slope of the line tangent to $f(x) = -x^{2}-3x+5$ at $x = 2$ ?
Explanation: The slope of the tangent line is $ \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$ $ = \lim_{h \to 0} \frac{(-(x+h)^{2}-3(x+h)+5) - (-x^{2}-3x+5)}{h}$ $ = \lim_{h \to 0} \frac{(-(x^{2}+2x h+h^{2})-3(x+h)+5) - (-x^{2}-3x+5)}{h}$ $ = \lim_{h \to 0} \frac{-x^{2}-2(x h)-h^{2}-3x-3h+5+x^{2}+3x-5}{h}$ $ = \lim_{h \to 0} \frac{-2(x h)-h^{2}-3h}{h}$ $ = \lim_{h \to 0} -2x-h-3$ $ = -2x-3$ $ = (-2)(2)-3$ $ = -7$